Evaluate the improper integral if it exists. $\int_{0}^{8}\dfrac{1}{\sqrt[ 3]{x}}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac32$ (Choice B) B $6$ (Choice C) C $8$ (Choice D) D The improper integral diverges.
Solution: First, let's rewrite the improper integral: $\int_{0}^{8}\dfrac{1}{\sqrt[ 3]{x}}\,dx =\lim_{a\to0^+}\int_a^{8} \dfrac{1}{\sqrt[ 3]{x}}\,dx $ We can now evaluate the integral: ∫ 1 0 1 x √ d x = lim a → 0 + ∫ 8 a 1 x √ 3 d x = lim a → 0 + [ 3 2 x 2 3 ] 8 a = lim a → 0 + ⎛ ⎝ 6 − 3 2 ( a ) 2 3 ⎞ ⎠ = lim a → 0 + ( 6 ) − lim a → 0 + ⎛ ⎝ 3 2 ( a ) 2 3 ⎞ ⎠ = 6 − 0 = 6 \begin{aligned} \phantom{\int_0^{1}\dfrac1{\sqrt{x}}dx}&=\lim_{a\to0^+}\int_a^{8} \dfrac{1}{\sqrt[ 3]{x}}\,dx\\ \\ \\ &=\lim_{a\to0^+}\Big[\dfrac{3}{2}x\^{\frac{2}{3}}\Big]_a^8\\ \\ \\ &=\lim_{a\to0^+}\left(6-\dfrac{3}{2}(a)\^{\frac{2}{3}}\right)\\ \\ \\ &=\lim_{a\to0^+}\left(6\right)-\lim_{a\to0^+}\left(\dfrac{3}{2}(a)\^{\frac{2}{3}}\right)\\ \\ &=6-0\\ \\ &=6 \end{aligned} The answer: $\int_{0}^{8}\dfrac{1}{\sqrt[ 3]{x}}\,dx =6$